$\int \dfrac{4\sin(x)}{3+\cos(x)}\,dx\,= $ $+~C$
Notice that we can rewrite the integral as $4 \int \dfrac{1}{3+\cos(x)}\,\cdot \sin(x)\,dx\,$. If we let $ {u=3+\cos(x)}$, then ${du=-\sin(x) \, dx}$ and $-du=\sin(x) \, dx} $ Substituting gives us: $\begin{aligned}4 \int \dfrac{1}{{3+\cos(x)}}\,\cdot \sin(x)\,dx}\, &=4 \int \dfrac{1}{ u}\,(-du)}\\\\\\ &=-4 \int \dfrac{1}{\ u}\,du\end{aligned}$ We recognize this antiderivative. $\begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx}&= -4 \int \dfrac{1}{\ u}\,du\\\\\\ &=-4\ln|u|+C\\\\\\\end{aligned}$ We can now substitute back to find the antiderivative in terms of $x$. ∫ e x 1 + e 2 x d x = − 4 ln | u | + C = − 4 ln | 3 + cos ( x ) | + C = − 4 ln ( 3 + cos ( x ) ) + C ( 3 + cos ( x ) is always positive ) \begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx~}&=-4\ln|u|+C\\\\\\\ &=-4\ln|3+\cos(x)|+C\\\\\\ &=-4\ln(3+\cos(x))+C&& {\gray{(3+\cos(x) }}\gray{\text{ is always positive})}\end{aligned} The answer: $\int \dfrac{4\sin(x)}{3+\cos(x)}\,dx\,= -4\ln(3+\cos(x))+C$